JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020
Answers
A beaker of radius r is filed with water ( reflective index 4/3) upto a height H as shown in the figure on the left. The beaker is kept on a horizontal table rotating with angular speed ω. This makes the water surface curved so that the difference in the height of water level at the center and at the circumference of the beaker is h (h << H , h << r) as shown in figure on the right. Take this surface to be approximately spherical with the radius of curvature R. which is to the following is/are correct.
solution : see the diagram,
from ∆OQB,
R² = (R - h)² + r²
⇒R² = R² - 2Rh + h² + r²
⇒2hR = h² + r²
⇒R = (h² + r²)/2h .........(1)
now considering equation of surface is,
y = y₀ + h
here , h = v²/2g = (ωr)²/2g = ω²r²/2g ......(2)
Now using formula, μ₂/v - μ₁/u = (μ₂ - μ₁)/R
⇒1/v - (4/3)/-(H - h) = (1 - 4/3)/-R
⇒1/v + 4/3(H - h) = 1/3R
as h << H so, (H - h) ≈ H
⇒1/v = 1/3R - 4/3H
⇒1/v = 1/3(h² + r²)/2h - 4/3H [ from eq (1) ]
⇒1/v = 2h/3(h² + r²) - 4/3H
as h << r, so h² + r² ≈ r²
⇒1/v = 2h/3r² - 4/3H = -4/3H [ 1 - ω²H/4g] [ from equation (2) ]
⇒v = -3H/4 [ 1 + ω²H/g]¯¹ [ apply binomial expansion ]
so apparent depth of the bottom of the beaker is 3H/4 [ 1 + ω²H/4g]¯¹
Therefore option (A) and (D) is correct choices.
